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  "body": "\n## Proof\n\n1. Let $(X, \\tau)$ be a [[topological space|0CNQXVBHCX]] and let $S \\subseteq X$.\n2. Let $\\mathcal{F}$ be the family of all [[closed sets|0G5AV9Q4RR]] $F$ of $X$ with $S \\subseteq F$. By the definition of the [[closure|0DRKT1QJR7]], $\\overline{S}$ is the [[intersection|0839XNQXDK]] of the members of $\\mathcal{F}$.\n3. The family $\\mathcal{F}$ is nonempty. Indeed $X \\setminus X = \\varnothing$ is in $\\tau$, so $X$ is closed, and $S \\subseteq X$.\n4. We claim that $X \\setminus \\overline{S}$ is the [[union|05GK32H72H]] of the [[complements|0B1FSS1VZW]] $X \\setminus F$ for $F \\in \\mathcal{F}$.\n5. Let $x \\in X$. Then $x \\notin \\overline{S}$ if and only if $x \\notin F$ for some $F \\in \\mathcal{F}$, that is, $x \\in X \\setminus F$ for some $F \\in \\mathcal{F}$. This proves the claim.\n6. For each $F \\in \\mathcal{F}$, the set $X \\setminus F$ is [[open|01ZQQDCERP]], because $F$ is closed.\n7. By the second axiom of a topology, a union of members of $\\tau$ is in $\\tau$. By steps 4 and 6, $X \\setminus \\overline{S}$ is open.\n8. By the definition of a closed set, $\\overline{S}$ is closed in $X$.\n\n## Idea\n\nThe closure is an intersection of closed sets. Its complement is the union of their complements, which are open. A union of open sets is open by the axioms, so the closure is closed.\n",
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